If h(x)=∑n=1∞xn+1nh(x) = \sum_{n=1}^{\infty} \frac{x^{n+1}}{n}h(x)=∑n=1∞nxn+1, what is h′(x)h'(x)h′(x) using term-by-term differentiation?
∑n=1∞xnn\sum_{n=1}^{\infty} \frac{x^n}{n}∑n=1∞nxn
∑n=1∞(n+1)xnn\sum_{n=1}^{\infty} \frac{(n+1)x^n}{n}∑n=1∞n(n+1)xn
∑n=0∞xn=11−x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}∑n=0∞xn=1−x1 for ∣x∣<1|x| < 1∣x∣<1
∑n=2∞xn−1n−1\sum_{n=2}^{\infty} \frac{x^{n-1}}{n-1}∑n=2∞n−1xn−1